find three consecutive odd integers such that the square of the first two exceeds the square of the third number by 15.

Question

Need help with this word problem. I thought it would be (x^2+2x)= (x+4)-15 but that clearly did not work out

Answer 1

Three consecutive odd integers: x, x+2, x+4

 

Your problem says "The square of the first two..."

That is very unclear.

Assuming you mean the sum of the squares of the first two...

 

(x+x+2)² = (x+4)²+15

(2x+2)² = x²+8x+16+15

4x² + 8x + 4 = x² + 8x + 31

3x² - 27 = 0

3(x²-9) = 0

x² - 9 = 0

(x-3)(x+3) = 0

 

x = 3 or x= -3

Your ntegers are:

3, 5, 7

or

-3, -1, 1

************************

 

If you mean the sum of the squares of the 

first two:

x² + (x+2)² = x² + 8x + 31

x² + x² + 4x + 4 = x² + 8x + 31

2x² + 4x + 4 = x² + 8x + 31

x² - 4x - 27 = 0

 

x = [4 ±√((-4)²-4(-27))]/2

x = [4 ±√124]/2

This will not be an integer so I assume the

first answer was correct.